Three Faraday’s of electricity are passed through molten Al2O3, aq. solution of CuSO4 and molten NaCl taken in three different electrolytic cells. The amounts of Al, Cu and Na deposited at the electrodes will be

Al3++ 3e-→Al 1 mol Al = 3F Cu2++ 2e-→Cu 1 mol Cu= 2F 3 F = 32mol Na++ e-→Na 1 mol = 1F 3 F = 3 mol Al : Cu : Na = 1 : 32:3 =1: 1.5 :3

Three Faraday’s of electricity are passed through molten Al2O3, aq. solution of CuSO4 and molten NaCl taken in three different electrolytic cells. The amounts of Al, Cu and Na deposited at the electrodes will be

Al3++ 3e-→Al 1 mol Al = 3F Cu2++ 2e-→Cu 1 mol Cu= 2F 3 F = 32mol Na++ e-→Na 1 mol = 1F 3 F = 3 mol Al : Cu : Na = 1 : 32:3 =1: 1.5 :3

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Three Faraday’s of electricity are passed through molten Al₂O₃, aq. solution of CuSO₄ and molten NaCl taken in three different electrolytic cells. The amounts of Al, Cu and Na deposited at the electrodes will be

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1 mole : 1.5 mole : 2 moles

3 moles : 2 moles : 1 mole

1 mole : 2 moles : 3 moles

1 mole : 1.5 mole : 3 moles

answer is B.

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$A l^{3 +} + 3 e^{-} \to A l 1 m o l A l = 3 F C u^{2 +} + 2 e^{-} \to C u 1 m o l C u = 2 F 3 F = \frac{3}{2} m o l N a^{+} + e^{-} \to N a 1 m o l = 1 F 3 F = 3 m o l A l : C u : N a = 1 : \frac{3}{2} : 3 = 1 : 1.5 : 3$

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Three Faraday’s of electricity are passed through molten Al2O3, aq. solution of CuSO4 and molten NaCl taken in three different electrolytic cells. The amounts of Al, Cu and Na deposited at the electrodes will be

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Three Faraday’s of electricity are passed through molten Al₂O₃, aq. solution of CuSO₄ and molten NaCl taken in three different electrolytic cells. The amounts of Al, Cu and Na deposited at the electrodes will be